Problem: What is the extraneous solution to these equations? $\dfrac{x^2 + 3}{x + 2} = \dfrac{3x + 13}{x + 2}$
Explanation: Multiply both sides by $x + 2$ $ \dfrac{x^2 + 3}{x + 2} (x + 2) = \dfrac{3x + 13}{x + 2} (x + 2)$ $ x^2 + 3 = 3x + 13$ Subtract $3x + 13$ from both sides: $ x^2 + 3 - (3x + 13) = 3x + 13 - (3x + 13)$ $ x^2 + 3 - 3x - 13 = 0$ $ x^2 - 10 - 3x = 0$ Factor the expression: $ (x + 2)(x - 5) = 0$ Therefore $x = -2$ or $x = 5$ At $x = -2$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -2$, it is an extraneous solution.